V=IR...i am confused..plz someone explain...by the equation...?

Question

ok, we all knoe ohm's law: V=IR..if r increases,provided I remains the same,V will ALSO increase....but that does happen!....V decreases...i think i am BADLY confused in electricity...plz plz plz help! i am getting confusions after confusions....oh plz someone clarify!

V IS GOING DOWN...REFER TO THIS
http://in.answers.yahoo.com/question/index;_ylt=AiVtN9rPqpyAhQUyJMPYidiQHQx.?qid=20070217031133AAJcalH

WHY IS V GOING DOWN HERE THEN?

Answer 1

http://micro.magnet.fsu.edu/electromag/java/ohmslaw/

Mathematically Ohm's law can be written as:

I = E/R

where I is current in amperes, E is the applied voltage in volts, and R is resistance in ohms.

This circuit contains an ammeter measuring current flow in milliamperes. Notice that as you increase voltage, current flow increases. As you increase resistance, current flow decreases.

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It is important to note that resistance cannot be changed by changing voltage or current. Resistance in a circuit is a physical constant. Resistance in a circuit can only be changed by changing components or resistors rated at more or fewer ohms. The changing of resistance in this circuit simulates the physical changing of resistors with different ratings.
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From wikipedia:
Ohm's law applies to conductors whose resistance is (substantially) independent of the applied voltage (or equivalently the injected current). That is, Ohm's law only applies to the linear portion of the I vs. V curve centered around the origin. The equation is just too simple to encompass devices described by a more complicated I vs. V relationship.

So, always keep the resistance constant, and then study the relationship between current and voltage. not between voltage and resistance.

Answer 2

In a real life situation, say for example in the house Voltage is fixed. A bulb has an element that has a very high resistance but uses very low current. On the other hand a toaster has an element with a relatively low resistance bu uses more current.

To verify check the values on these items, to measure the amount of actual current going through you would need a clamp on ampere meter.

The following web pages should provide you more information. Try a few sample inputs into the different variables V, R, I etc and see the magic come to life, yourself:




http://www.the12volt.com/ohm/ohmslaw.asp

Answer 3

The problem with the equation V=IR is that it is poorly written to reflect the observable, intuitive, cause-and-effect universe, just like when people write Newton's second law as F=ma. It is easier to understand if you write it as I = V/R because usually there is a given amount of potential difference (V) that causes current to flow. (Just like Newton's second law, a=F/m, usually there is a force that causes acceleration not vice-versa).

It is not normal to "make I remain the same" when you change R. The only way to do that would be to use a power source that automatically compensates for changing resistance in the circuit by increasing its "EMF" (or voltage output).

Most power sources, such as chemical batteries, have a constant EMF (voltage output). For example a 9 volt battery. That means the total voltage DROP across all circuit elements (all resistances) must equal the EMF. SImple to understand: total increase in voltage in a circuit caused by power sources must equal total voltage drop.

The problem is that if you increase any resistance in the circuit, but keep the same EMF of the power supply, the current drops. See, I=V/R.

I suspect that when you are increasing R in your circuit, your current is not being kept constant. Instead, I is decreasing.

Here is another way to look at your experiment: use 2 resistances in series, one much larger than the other so the current will not change much if you increase the resistance of the smaller one. Example: 10,000 ohm fixed resistance, in series with a variable resistor (100 ohms to 1000 ohms for example). As you increase the variable resistor, while measuring only the voltage drop across that resistor, you should measure increasing voltage drop across the variable resistor. That is because the current should not change much.

Answer 4

Ohm's law states that, in an electrical circuit, the current passing through a conductor, from one terminal point on the conductor to another terminal point on the conductor, is directly proportional to the potential difference (i.e. voltage drop or voltage) across the two terminal points and inversely proportional to the resistance of the conductor between the two terminal points.

For real devices (resistors, in particular), this law is usually valid over a large range of values of current and voltage, but exceeding certain limitations may result in losing simple direct proportionality (e.g. temperature effects, see below).


A voltage source, V, drives an electric current, I , through resistor, R, the three quantities obeying Ohm's law: V = IR.In mathematical terms, this is written as:

Simply apply this and u will find the desirable result.

Answer 5

you need not take so much tension about that let me simply explain u:
ohms law states that the ratio of (pot. diff/current) yields us a constant called resistance. on increasing the resistance(r)
ie;(v/i)=r
the flow of current across the two ends will be lowered and hence give greater potential difference(v). u can experimentally prove it assemble all the components of the ohm's circuit, now take out the key so that current flows in the circuit, now change the values of resistance thru rheostat, note the change in current by using ammeter and take atleast 2-3 readings for
perfection.substitute in the formula and get the values .U will achieve the theoritical result which i have discussed earlier.

Answer 6

V=IR is the mathemaical equation to represent Ohm's law. V represents the potential difference across the two ends of the conductor. I represents the current flowing in the circuit. And R is the resistance constant. Actually it is that V is directly proportional to I at constant temperature. If you need any further assistance, mail me.

Answer 7

First I find it easier to rewrite anything that isnt grouped together, in this case, the water. MgCO3 + HCl --> MgCl2 + H2O + CO2 First lets start with balancing the hydrogens and see where we stand. There are 2 hydrogens on the right side, and one of the left... so lets make it two on the left. MgCO3 + 2HCl --> MgCl2 + H2O + CO2 Lets see if anything is not balanced. Left side / Right Side Mg: 1/1 C: 1/1 O: 3/3 H: 2/2 Cl: 2/2 Looks like we are all balanced, so your final, balanced equation is: MgCO3 + 2HCl --> MgCl2 + H2O (or HOH, same thing) + CO2 *Sorry for the delay in answering your question, YA has been down

Answer 8

One way of looking at it is that the voltage is the force that drives current through the resistance. If resistance to current flow increases then current flow will decrease unless the force is increased. So if R increases then V must be increased if I is to stay the same.

Answer 9

As resistance increases, voltage flowing over the circuit will decrease. Think of all of this in terms of water, resistance would be the size of the pipe. To increase the resistance, you would decrease the size of the pipe, that would allow less water to flow through the pipe. I would be the pressure pushing the water through. It would remain the same.

Answer 10

Imagine the two resistances in series but must having different values.( as you want to keep the current constant ). If the current will remain same,for higher resistance the value of potential difference across it will be also more or increased as compared with the other resistance. hence if u increase resistance while keeping the current same, definately potential difference across it will increase.